0
$\begingroup$

I went back to many references in order to fix some of the confusions that I have on many concepts in signal spectral representation. I concluded that:

1) Deterministic signals may be represented with the ESD if it is about an energy signal (ESD is the Fourier transform of the energy signal ACF), or (not both) the PSD if it is about a power signal (PSD is the Fourier transform of the power signal ACF). Where, for a deterministic signal $x(t)$, the energy signal ACF is expressed as

$\varphi_x(\tau)=\int_{-\infty}^{\infty}x(t)x^*(t-\tau)\text{d}t$,

and the power signal ACF is expressed as

$R_x(\tau)=\lim_{T\to\infty}\frac{1}{T}\int_{-T/2}^{T/2}x(t)x^*(t-\tau)\text{d}t$,

or $R_x(\tau)=\frac{1}{T}\int_{t_0}^{t_0+T}x(t)x^*(t-\tau)\text{d}t$ for a signal of period $T$.

2) Random WSS signals (regardless ergodicity) are power signals, so they are represented by their PSD. The ACF of a WSS random signal $x(t)$ is expressed as

$R_x(\tau)=\mathbb{E}\{x(t)x^*(t-\tau)\}$.

3) Signals neither energy or power have no spectral representation.

4) $\int_{-\infty}^{\infty}|x(t)|\text{d}t<\infty$ is a necessary condition for the existence of the Fourier transform of $x(t)$, but $\int_{-\infty}^{\infty}|x(t)|^2\text{d}t<\infty$ is sufficient.

5) If Fourier transform doesn't exist, neither ESD nor PSD exist.

6) Some references say that the Fourier transform (FT) didn't exist for periodic signals, while the FT of sinusoidal signals can be easily derived using Euler formula. What's wrong here?

7) Parseval's identity (sometimes referred as Rayleigh energy theorem) is applicable only for energy signals.

Could you please check for the rightness of the above items?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.